Question

6 sur 32 lors dune neutralisation, 30 ml dune solution de h₂so₄ ont neutralisé 45 ml dune solution de naoh à une concentration de 0,2 mol/l. quelle est la concentration molaire de la solution dacide sulfurique (h₂so₄) ? h₂so₄ + 2 naoh → na₂so₄ + 2 h₂o 5 points a 2,5 mol/l b 0,1 mol/l c 0,75 mol/l d 3,2 mol/l e 0,15 mol/l

Explanation:

Step1: Recall the neutralization reaction stoichiometry

The reaction is $\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O}$. So the mole ratio of $\ce{H2SO4}$ to $\ce{NaOH}$ is $1:2$.

Step2: Calculate moles of $\ce{NaOH}$

Molarity formula: $n = C\times V$ (where $n$ is moles, $C$ is concentration, $V$ is volume in liters).
Volume of $\ce{NaOH}$: $V_{NaOH} = 45\space mL = 0.045\space L$, $C_{NaOH} = 0.2\space mol/L$.
$n_{NaOH} = 0.2\space mol/L\times0.045\space L = 0.009\space mol$.

Step3: Find moles of $\ce{H2SO4}$ from mole ratio

From the reaction, $n_{\ce{H2SO4}}=\frac{n_{\ce{NaOH}}}{2}$.
$n_{\ce{H2SO4}}=\frac{0.009\space mol}{2}= 0.0045\space mol$.

Step4: Calculate concentration of $\ce{H2SO4}$

Volume of $\ce{H2SO4}$: $V_{\ce{H2SO4}} = 30\space mL = 0.03\space L$.
Molarity $C_{\ce{H2SO4}}=\frac{n_{\ce{H2SO4}}}{V_{\ce{H2SO4}}}=\frac{0.0045\space mol}{0.03\space L}= 0.15\space mol/L$? Wait, no, wait: Wait, let's recheck. Wait, $n_{NaOH}=0.2*0.045 = 0.009$. Then $n_{H2SO4}=n_{NaOH}/2 = 0.0045$. Then $C = 0.0045 / 0.03 = 0.15$? Wait, but let's check again. Wait, 0.0045 divided by 0.03: 0.0045 / 0.03 = 0.15? Wait, no, 0.03 L is 30 mL. 0.0045 mol / 0.03 L = 0.15 mol/L? Wait, but let's check the options. Option E is 0.15 mol/L. Wait, but let's re - calculate:

Wait, $V_{H2SO4}=30\space mL = 0.03\space L$, $V_{NaOH}=45\space mL = 0.045\space L$, $C_{NaOH}=0.2\space mol/L$.

From the reaction: $n_{H2SO4}/n_{NaOH}=1/2$ => $n_{H2SO4}=n_{NaOH}/2$.

$n_{NaOH}=C_{NaOH}\times V_{NaOH}=0.2\space mol/L\times0.045\space L = 0.009\space mol$.

$n_{H2SO4}=0.009\space mol / 2 = 0.0045\space mol$.

$C_{H2SO4}=n_{H2SO4}/V_{H2SO4}=0.0045\space mol / 0.03\space L = 0.15\space mol/L$.

Wait, but let's check the options. Option E is 0.15 mol/L. But wait, maybe I made a mistake? Wait, no, let's check the calculation again. 0.2 * 0.045 = 0.009. 0.009 / 2 = 0.0045. 0.0045 / 0.03 = 0.15. Yes. So the concentration of $\ce{H2SO4}$ is 0.15 mol/L.

Answer:

E. 0,15 mol/L